Independent and Dependent Events

One of the most important concepts in probability is that of independent events

Two events E and F are independent if the occurrence of event E does not affect the probability of event F.

If the occurrence or non-occurrence of E₁ does not affect the probability of occurrence of E₂, then

P(E₂ | E₁) = P(E₂)

and E₁ and E₂ are said to be independent events.

Otherwise they are said to be dependent events.

[Recall from Conditional Probability that the notation P(E₂ | E₁) means "the probability of the event E₂ given that E₁ has already occurred".]

Two Events

Let's consider "E₁ and E₂" as the event that "both E₁ and E₂ occur".

If E₁ and E₂ are dependent events, then:

P(E₁ and E₂) = P(E₁) × P(E₂ | E₁)

If E₁ and E₂ are independent events, then:

P(E₁ and E₂) = P(E₁) × P(E₂)

Three Events

For three dependent events E₁, E₂, E₃, we have

P(E₁ and E₂ and E₃) = P(E₁) × P(E₂ | E₁) × P(E₃ | E₁ and E₂)

For three independent events E₁, E₂, E₃, we have

P(E₁ and E₂ and E₃) = P(E₁) × P(E₂) × P(E₃)

Example 1

If the probability that person A will be alive in $\displaystyle{20}$ years is $\displaystyle{0.7}$ and the probability that person B will be alive in $\displaystyle{20}$ years is $\displaystyle{0.5}$, what is the probability that they will both be alive in $\displaystyle{20}$

$20$ years?

ANSWER

 These are independent events, so

P(E₁ and E₂) = P(E₁) × P(E₂) = 0.7 × 0.5 = 0.35

[Note, however, that if person A knows person B, then they will be dependentevents, especially if A is married to B.]

Example 2

A fair die is tossed twice. Find the probability of getting a $\displaystyle{4}$ or $\displaystyle{5}$ on the first toss and a $\displaystyle{1}$, $\displaystyle{2}$, or $\displaystyle{3}$ in the second toss.

ANSWER

P(E₁) = P(4 or 5) = $\displaystyle\frac{2}{{6}}=\frac{1}{{3}}$

P(E₂) = P(1, 2 or 3) $\displaystyle=\frac{3}{{6}}=\frac{1}{{2}}$

They are independent events, so

$\displaystyle{P}{\left({E}_{{1}}{\quad\text{and}\quad}{E}_{{2}}\right)}$ $\displaystyle={P}{\left({E}_{{1}}\right)}\times{P}{\left({E}_{{2}}\right)}$ $\displaystyle=\frac{1}{{3}}\times\frac{1}{{2}}$ $\displaystyle=\frac{1}{{6}}$

Example 3

Two balls are drawn successively without replacement from a box which contains $\displaystyle{4}$white balls and $\displaystyle{3}$ red balls. Find the probability that

(a) the first ball drawn is white and the second is red;

(b) both balls are red.

ANSWER

(a) The second event is dependent on the first.

P(E₁) = P(white) = $\displaystyle\frac{4}{{7}}$

There are 6 balls left and out of those 6, three of them are red. So the probability that the second one is red is given by:

P(E₂ | E₁) = P(red) $\displaystyle=\frac{3}{{6}}=\frac{1}{{2}}$

Dependent events, so

$\displaystyle{P}{\left({E}_{{1}}\ \text{and}\ {E}_{{2}}\right)}={P}{\left({E}_{{1}}\right)}\times{P}{\left({E}_{{2}}{|}{E}_{{1}}\right)}=\frac{4}{{7}}\times\frac{1}{{2}}=\frac{2}{{7}}$

(b) Also dependent events. Using similar reasoning, but realising there will be 2 red balls on the second draw, we have:

$\displaystyle{P}{\left({R}{R}\right)}=\frac{3}{{7}}\times\frac{2}{{6}}=\frac{1}{{7}}$